[SOLVED] Showing that an expression is divisible by 7

**mark1950** · June 12th 2009, 11:23 PM

Show that $2^n + 2^{n+1} + 2^{n+2}$ is divisible by 7 where n is a positive integer. How do I do this question? I solved it by just showing like

$2^1 + 2^{1+1} + 2^{1+2}$
which equals to 14 and is divisible by 7

$2^2 + 2^{2+1} + 2^{2+2}$
which equals to 28 and is divisible by 7

hence, proven.

But is it correct to show it just like that?

**flyingsquirrel** · June 13th 2009, 12:17 AM

Originally Posted by mark1950

$2^1 + 2^{1+1} + 2^{1+2}$
which equals to 14 and is divisible by 7

Yes. Another way to see this is to factor :

$2^1+2^2+2^3=2^1(2^0+2^1+2^2)=2\times 7.$

Originally Posted by mark1950

$2^2 + 2^{2+1} + 2^{2+2}$
which equals to 28 and is divisible by 7

Similarly,

$2^2+2^3+2^4=2^2(2^0+2^1+2^2)=2^2\times 7.$

Using this idea, can you show that 7 divides $2^n+2^{n+1}+2^{n+2}$ ?

Originally Posted by mark1950

But is it correct to show it just like that?

You only showed that 7 divides $2^n+2^{n+1}+2^{n+2}$ if $n=1$ or 2...

**mark1950** · June 13th 2009, 01:27 AM

Show that 2^n + 2^{n+1} + 2^{n+2} is divisible by 7 where n is a positive integer.

Hmm...flying squirrel, how would you do it if I just gave you the question?

**mr fantastic** · June 13th 2009, 01:47 AM

Originally Posted by mark1950

Show that $2^n + 2^{n+1} + 2^{n+2}$ is divisible by 7 where n is a positive integer. How do I do this question? I solved it by just showing like

$2^1 + 2^{1+1} + 2^{1+2}$
which equals to 14 and is divisible by 7

$2^2 + 2^{2+1} + 2^{2+2}$
which equals to 28 and is divisible by 7

hence, proven.

But is it correct to show it just like that?

$2^n + 2^{n+1} + 2^{n+2} = 2^n (1 + 2 + 2^2)$ ....

**mark1950** · June 13th 2009, 02:13 AM

$2^n + 2^{n+1} + 2^{n+2} = 2^n (1 + 2 + 2^2)$ ....

Is that the end answer? I don't know but it looks incomplete to me. Do you mean that if I just write that then I have shown that $2^n + 2^{n+1} + 2^{n+2}$ is divisible by 7 where n is a positive integer?

**yeongil** · June 13th 2009, 02:21 AM

Originally Posted by mark1950

Is that the end answer? I don't know but it looks incomplete to me. Do you mean that if I just write that then I have shown that $2^n + 2^{n+1} + 2^{n+2}$ is divisible by 7 where n is a positive integer?

No. I think mr fantastic wanted you to simplify the expression inside the parentheses...

01

**mark1950** · June 13th 2009, 02:44 AM

Originally Posted by yeongil

No. I think mr fantastic wanted you to simplify the expression inside the parentheses...

01

Hey...let's get back to the real question. In other words, I don't really get what flyingsquirrel is trying to say because he ends up with another question. I just wanted to know if my solution above in the description of my question is correct or not because as flyingsquirrel said, I was just showing that 7 divide by $2^n + 2^{n+1} + 2^{n+2}$ is true for n = 1 and n = 2.

Can my solution be accepted by examiners?

which equals to 14 and is divisible by 7

which equals to 28 and is divisible by 7

hence, proven.

I would kindly prefer just a simple yes if my solution is right and if my solution is wrong, please do kindly explain...

remember to include yes or no! Thanks.

**Chop Suey** · June 13th 2009, 03:24 AM

You only showed that it is true for n = 1 and n = 2. You did not show it to be true for all positive integers. So in short, your solution is not complete.

Pay attention to flyingsquirrel's and mrfantastic's hints and use it to factor the general expression.

**mark1950** · June 13th 2009, 07:18 AM

Originally Posted by Chop Suey

You only showed that it is true for n = 1 and n = 2. You did not show it to be true for all positive integers. So in short, your solution is not complete.

Pay attention to flyingsquirrel's and mrfantastic's hints and use it to factor the general expression.

Factorise the expression? You mean I have to find all the values of n? But the values would be until infinity...and until what number of n should I list until its enough to convince the examiners that

is divisible by 7 where n is a positive integer?

$2^1+2^2+2^3=2^1(2^0+2^1+2^2)=2 \times 7. $
$2^1+2^2+2^3=2^2(2^0+2^1+2^2)=2^2 \times 7. $
$2^1+2^2+2^3=2^3(2^0+2^1+2^2)=2^3 \times 7.$
$2^1+2^2+2^3=2^4(2^0+2^1+2^2)=2^4 \times 7.$

and on and on?

**flyingsquirrel** · June 13th 2009, 08:26 AM

Originally Posted by mark1950

Factorise the expression? You mean I have to find all the values of n?

No. You just have to write $2^n+2^{n+1}+2^{n+2}$ as the product of two numbers. (see mr fantastic's post)

**yeongil** · June 13th 2009, 02:44 PM

Look again at what mr fantastic wrote:

$2^n + 2^{n+1} + 2^{n+2} = 2^n (1 + 2 + 2^2)$

Just simplify the expression inside the parentheses, and you are done. Can you take the above and simplify it for us, please?

01

**mark1950** · June 13th 2009, 11:32 PM

$2^n + 2^{n+1} + 2^{n+2} = 2^n (1 + 2 + 2^2)$
$= 2^n \times 7$

End? Like this?

**mr fantastic** · June 14th 2009, 02:16 AM

Originally Posted by mark1950

$2^n + 2^{n+1} + 2^{n+2} = 2^n (1 + 2 + 2^2)$
$= 2^n \times 7$

End? Like this?

Well, if you were asked to explain why $2^n \times 7$ is divisible by 7, what would you say ....?

**mathaddict** · June 14th 2009, 02:23 AM

Originally Posted by mark1950

$2^n + 2^{n+1} + 2^{n+2} = 2^n (1 + 2 + 2^2)$
$= 2^n \times 7$

End? Like this?

Yes .. If you divide (2^n)7 by 7 , it leaves you with 2^n regardless of the value of the positive integer ,n .

**Showcase_22** · June 14th 2009, 03:39 AM

I don't think anyone has mentioned it yet but a proof by induction works pretty well.

Firstly, as you have shown, it is true for n=1.

Assume true for n=k:

$ 2^k + 2^{k+1} + 2^{k+2} $

For n=k+1:

$ 2^{k+1} + 2^{k+2} + 2^{k+3} $

and using Mr Fantastic's factorisation: $2(2^k+2^{k+1}+2^{k+2})$ which is divisible by 7 by our induction hypothesis.

(alternatively $ 2^{k+1} + 2^{k+2} + 2^{k+3}=2^k(2+2^2+2^3)=2^k \cdot 14$ which is divisible by 7).

Hence it's true $\forall n \in \mathbb{N}$ .

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